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This question is an extension of this prior one, focussing on the conditions under which a solution to that prior problem is possible. That prior problem considered the case of three points in three dimensions, each far from each other and not approximately colinear, and the number of planes that were a unit distance from all three points. In brief, the answer is $2^3 = 8$, corresponding to planes tangent to the "top" of all three unit spheres centered on the points; on the "bottom" of all three unit spheres; on the "top" of sphere $A$ and the "bottom" of $B$ and $C$, and so forth. (A figure in one solution of the problem may help you visualize what is going on.)

The current question centers on the constraints on the locations of the three points in three dimensions that admit this solution of eight planes each of which is a unit distance from the three points. Clearly, if the points are too close together (e.g., distance less than 2), the above solution cannot hold, as one cannot find planes that are tangent to all configurations of "top" and "bottom" for the nearby spheres. But what are the full sets of conditions that admit the eight planes?

Without loss of generality, assume the three points in three dimensions are constrained to the $(x,y)$ plane: point $A$ at the origin $(0,0,0)$, point $B$ on the $x$ axis $(x_B,0,0)$, and point $C$ in the upper half plane $(x_C,y_C,0)$ where $y_C \ge 0$.

What relations among the variables admit all eight planes? What relations among the variables admit exactly four planes? (Note that if the points are co-linear--i.e., $y_C=0$--then there is an infinite number of planes.)

Addendum

@Rahul states that sphere $A$ must lie outside the cones defined by spheres $B$ and $C$ for all eight tangent planes to exist. This is not sufficient, as this figure illustrates:

enter image description here

The blue sphere lies outside the cone(s) defined by the green and red, yet one cannot obtain all eight tangent planes.

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David G. Stork
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  • For a plane to exist that goes "above" sphere $A$ and "below" spheres $B$ and $C$ or vice versa, sphere $A$ must lie entirely outside the cone determined by spheres $B$ and $C$. –  Dec 19 '16 at 17:12
  • Rahul: The solution requires all the "above" and "below" conditions on the permuted points. So what, conditions are required there? – David G. Stork Dec 19 '16 at 17:18
  • I believe the conjunction of the condition in my comment and its two other permutations guarantees the existence of all eight planes, but I don't have a proof yet. –  Dec 19 '16 at 18:35
  • State the conditions as a test on the variables. For instance, if $x_B = 3$, $x_C = 1$ and $y_C = 2$ do we get eight planes? – David G. Stork Dec 19 '16 at 18:42
  • Oh, I see what you mean. Given that the radii are equal the cones are actually cylinders, and the condition reduces to checking that point $A$ is at distance greater than $2$ from the line joining $B$ and $C$ and so on, which has a standard formula. In your example, the point $C$ is at distance exactly $2$ from line $AB$, so two planes coincide (analogous to the inner tangent(s) of two circles that are themselves touching) and we only have seven distinct planes. –  Dec 19 '16 at 18:52
  • Sorry, in this case I meant the cylinder defined by the spheres. I was thinking of the more general case with unequal radii, where the cylinder would become a cone with vertex not between the two spheres. –  Dec 19 '16 at 21:11
  • Even the constraints of a cylinder simply do not suffice, as I'll let you convince yourself with a simple drawing. – David G. Stork Dec 19 '16 at 21:17
  • Sorry, I'm afraid I am unable to convince myself. –  Dec 19 '16 at 21:29

1 Answers1

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In my original answer I overcomplicated the problem by generalizing it to spheres of varying radii. For the case of unit radii the solution is much simpler. As in the question, we take the three points $A,B,C$ to lie in the $xy$-plane, with the $x$-axis along $AB$. Consider the three altitudes of the triangle $ABC$.

$$\begin{align} \text{number of tangent planes} &= 2 \\ &+ 2 \times (\text{number of altitudes of length > 2}) \\ &+ (\text{number of altitudes of length 2}) \end{align}$$

The first term counts the two planes $z=\pm 1$; these are always tangent to the three spheres.

For the second and third terms, consider the altitude from $C$, whose length is $|y_C|$. In the $yz$-plane, the spheres around $A$ and $B$ both project to the unit circle centered at $(0,0)$, and the sphere around $C$ projects to a unit circle centered at $(y_C,0)$. The two circles are disjoint, tangent, or intersecting depending on whether $|y_C|>2$, $|y_C|=2$, or $|y_C|<2$, in which case they have two, one, or zero internal tangents respectively (excluding the external tangents $z=\pm 1$). These correspond to two, one, or zero planes tangent to all three spheres.

In particular, if the three spheres are very close to each other, they form a triangle whose side lengths are nearly $2$, and there are no altitudes of length $\ge 2$. Thus we correctly predict only $2$ tangent planes, namely the ones entirely above or below the spheres.

  • If I understand your point 1, I am fairly sure it is false, as the computer graphics figure I generated above illustrates (if you imagine the cylinder). – David G. Stork Dec 19 '16 at 23:09
  • I did imagine the cylinder and do not see how it contradicts point 1 or point 2. Projected to the plane containing the three centers, we have a circle that meets the slab but neither lies entirely inside it nor is big enough to reach across to the other side, so the top and bottom tangent planes exist. On the other hand, since the circle does intersect the slab, two of the other tangent planes do not exist. If I am wrong, please let me know the coordinates of the spheres. // By the way, I updated my answer with a proof for the equal-radius case. –  Dec 19 '16 at 23:16
  • You must have made a mistake, then, when you wrote "$C_3$ does not lie entirely inside $K_{12}$. If it lies partially inside, you have problems. Right? – David G. Stork Dec 19 '16 at 23:21
  • No. I wrote, "The top and bottom tangent planes exist if and only if (i) $C_3$ does not lie entirely inside $K_{12}$, and (ii) $C_3$ does not meet both sides of the complement of $K_{12}$." As far as I can tell from your figure, $C_3$ does not lie entirely inside $K_{12}$. Also, $C_3$ does not meet both sides of the complement of $K_{12}$. Therefore, the right-hand side of the iff is true, so the left-hand side should be too, and it is: there are two tangent planes that all spheres are on the same side of (though they are "near" and "far" rather than "top" and "bottom" in your figure). –  Dec 19 '16 at 23:30
  • Stick with $r=1$ to simplify things, and focus on the problem and its essence. Now take three oranges or other spheres and place them on a table mutually touching (in an equilateral triangle). Now separate them from each other just slightly so they are no longer touching. Use a piece of paper as a plane and see if you can place it in eight tangent ways. You cannot. I believe your above statement says you can. – David G. Stork Dec 19 '16 at 23:51
  • Then you are misunderstanding my statements. Please re-read my first comment, where I literally state why "two of the other tangent planes do not exist". –  Dec 19 '16 at 23:57
  • I admit that my answer was overcomplicated and hard to understand. I have significantly simplified it now; let me know what you think. –  Dec 20 '16 at 02:00
  • Much better (and not needlessly complicating). +1 and $\surd$. – David G. Stork Dec 20 '16 at 02:14